題目

Given an m x n 2D binary grid grid which represents a map of '1's (land) and '0's (water), return the number of islands.

An island is surrounded by water and is formed by connecting adjacent lands horizontally or vertically. You may assume all four edges of the grid are all surrounded by water.

 Example 1: 
Input: grid = [
  ["1","1","1","1","0"],
  ["1","1","0","1","0"],
  ["1","1","0","0","0"],
  ["0","0","0","0","0"]
]
Output: 1

 Example 2: 
Input: grid = [
  ["1","1","0","0","0"],
  ["1","1","0","0","0"],
  ["0","0","1","0","0"],
  ["0","0","0","1","1"]
]
Output: 3
 Constraints: 
 m == grid.length 
 n == grid[i].length 
 1 <= m, n <= 300 
 grid[i][j] is '0' or '1'. 
 

方法

深度優先遍歷法

• 時間複雜度：O（）
• 空間複雜度：O（）

class Solution {
    public int numIslands(char[][] grid) {
        int row = grid.length, col = grid[0].length;
        int count = 0;
        for(int i=0;i<row;i++){
            for(int j=0;j<col;j++){
                if(grid[i][j]=='1'){
                    dfs(grid,i,j);
                    count++;
                }
            }
        }
        return count;
    }
    private void dfs(char[][] grid,int i,int j){
        int row = grid.length,col = grid[0].length;
        if(i>=row||i<0||j>=col||j<0||grid[i][j]=='0'){
            return ;
        }
        grid[i][j] = '0';
        dfs(grid,i+1,j);
        dfs(grid,i-1,j);
        dfs(grid,i,j-1);
        dfs(grid,i,j+1);
    }
}
 

並查集

• 時間複雜度：O（）
• 空間複雜度：O（）